How Does Current Divide in a Circuit with Multiple Resistors and Batteries?

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The discussion focuses on calculating the current in a circuit with three resistors (R1 = 2.40 Ω, R2 = 4.45 Ω, R3 = 6.10 Ω) and two batteries. Participants emphasize the application of Kirchhoff's Laws to determine current flow, specifically addressing the relationship between currents I1, I2, and I3. The confusion arises from the interpretation of current splitting and potential loss across resistors. Clear guidance is provided to simplify the problem by correctly applying Kirchhoff's Laws without unnecessary complications.

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Find the magnitude of the current in each resistor.
R1 = 2.40 Ω, R2 = 4.45 Ω, and R3 = 6.10 Ω.

I keep on attempting to use Kirchoff's Rules but it isn't working out. It doesn't click in my head if the current is flowing from the left battery, does the current still split? Is R3 double resistance in order to bring the current down to I?

Assuming there was another imaginary point right after the left terminal but before R2 and another point at a, I used the equation Va-Vb=(emf)-IR but since I don't know the Potential loss nor the current, I can't solve for either. Also placed more imaginary points mirrored to the right battery and set the Vb from both equations equal to each other but only thing I have left is 0=24V-4.45I1-6.85I2.. Grrr... Anyone give me a hint please?
 
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Forget those imaginary points and double resistance. This problem is not at all complicated if you use Kirchhoff' laws.

Let be the current flowing out from the left battery I2 and the current flowing out from the right battery be I1. You have a third current I3 which flows from a to b through R3. How is I3 related to I1 and I2?



ehild
 

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